Question

嘿伙计们，我完成了这个基本的C程序，它向输入任何给定数字组的用户显示有序集，最小值，最大值，平均值和中位数。我遇到的问题是我必须使用类似＆＃34; 3.2％f＆＃34;当我打印数字时，如何只打印每个数字所需的最小小数位数？例如，如果我输入5,5.5，-0.2313和4，我的程序将按顺序显示为-0.23,4.00,5.00和5.50。我怎样才能读到-2313,4,5和5.5？在此先感谢您的帮助。

#include <stdio.h>

int findSize();
void sortArray(int size, double num[]);
double findAverage(int size, double num[]);
double findMedian(int size, double num[]);
void findLowtoHigh(int size, double num[]);
void findHightoLow(int size, double num[]);

int main()
{
 while(1)
 {
  int size = findSize();

  if(size <= 1)
  {
   return 0;
  }

  double num[size];
  double lowest = 0;
  double highest = 0;
  double average = 0;
  double median = 0;

  fprintf(stdout, "\n"); 

  sortArray(size, num);  
  findLowtoHigh(size, num);   
  findHightoLow(size, num); 
  average = findAverage(size, num);  
  median = findMedian(size, num);

  fprintf(stdout, "\n\nLowest Value: %3.4f", num[0]);
  fprintf(stdout, "\nHighest Value: %3.4f", num[size-1]);
  fprintf(stdout, "\n\nAverage Value: %3.4f\n", average);  
  fprintf(stdout, "Median Value: %3.4f", median);

  fprintf(stdout, "\n");
 }

}

int findSize()
{
 int size;

 fprintf(stdout, "\nPlease enter size of the array: ");
 scanf("%d", &size);

 return size;
}

void sortArray(int size, double num[])
{
 for(int i = 0; i <= size - 1; i++)
 {
  int j = i+1;
  fprintf(stdout, "Please enter number %d: ", j);
  fscanf(stdin, "%lf", &num[i]);
 }

 if(size > 1)
 {
  double holder = 0;

  for(int y = 0; y < size - 1; y++)
  { 
   for(int k = 0; k < size - 1; k++)
   {
    if(num[k] > num[k+1])
    { 
     holder = num[k];
     num[k] = num[k+1];
     num[k+1] = holder;
    }
   }
  }
 }
}

void findLowtoHigh(int size, double num[])
{
 fprintf(stdout, "\nFrom least to greatest: ");

 for(int x = 0; x <= size - 1; x++)
 {
  fprintf(stdout, "%3.2f ", num[x]);
 }
}

void findHightoLow(int size, double num[])
{
 fprintf(stdout, "\nFrom greatest to least: ");

 int reverse = size - 1;

 while(reverse != -1)
 {
  fprintf(stdout, "%3.2f ", num[reverse]);
  reverse--;
 }
}

double findAverage(int size, double num[])
{
 double average = 0;

 for(int a = 0; a <= size - 1; a++)
 {
  average = average + num[a];
 }

 average = average / size;

 return average;
} 

double findMedian(int size, double num[])
{
 double median = 0;

 if(size % 2 == 0)
 {
  median = (num[size/2 - 1] + num[size/2])/2;
 }
 else
 {
  median = num[size/2];
 }

 return median;
}

Answer 1

如果没有自定义编码，很难得到你想要的一切，但“%g”至少可以处理你想要的大部分内容。领先的零是非常不可协商的，是全部。它通常选择最紧凑的符号表示法，包括在极端情况下的科学记数法。

Answer 2

让sprintf()执行繁重的工作，然后丢掉拖尾的零。

char *print_min_precision(char * buffer) {
  // look fo rthe decimal point, if not found return without altering buffer
  if (strchr(buffer, '.') == NULL)
    return buffer;
  // Find the last digit
  char *p = &buffer[strlen(buffer) - 1];
  // While not at the beginning and digit is zero ...
  while (p > buffer && *p == '0') {
    // Change '0' to `\0`
    *p-- = '\0';
  }
  if (p > buffer && *p == '.') {
    *p-- = '\0';
  }
  return buffer;
}


char buf[400]; // a BA buffer
// Print the number using some %f format.
// 6 here is the maximum number of fractional digits to use
sprintf(buf, "%.6f", num[reverse]);  
// Print the reduced number.
fprintf(stdout, "%s ", print_min_precision(buf));

通常，尝试以数字方式计算尾随小数具有许多边缘条件。最好打印到缓冲区并对其进行后处理。确保为以下内容提供足够的缓冲区：

print_min_precision(sprintf(buf, "%.6f", 10e300));

控制双精度

2 个答案: